Gọi S là tập hợp tất cả các giá trị thực của...

D

- Hướng dẫn giải

Đáp án D.

Xét hàm số $f\left(x\right)=\frac{x^2+m\text{x}+m}{x+1}$ trên $\left[1;2\right],$ có $y\text{'}=\frac{x^2+2\text{x}}{{\left(x+1\right)}^2}>0,\forall x\in \left[1;2\right]$

Suy ra

$\begin{array}{l}\underset{\left[1;2\right]}{\max }\left|f\left(x\right)\right|=\left\{\left|f\left(1\right)\right|;\left|f\left(2\right)\right|\right\}\\ =\left\{\left|\frac{2m+1}{2}\right|;\left|\frac{3m+4}{3}\right|\right\}\\ =\left\{\frac{\left|2m+1\right|}{2};\frac{\left|3m+4\right|}{3}\right\}\end{array}$

TH1. Với

$\begin{array}{l}\underset{\left[1;2\right]}{\max }\left|f\left(x\right)\right|=\left|\frac{2m+1}{2}\right|\\ =\left\{\begin{array}{l}\left|2m+1\right|=4\\ \frac{\left|2m+1\right|}{2}\ge \frac{\left|3m+4\right|}{3}\end{array}\right.\\ \iff m=-\frac{5}{2}.\end{array}$

TH2:

Với $\begin{array}{l}\underset{\left[1;2\right]}{\max }\left|f\left(x\right)\right|=\left|\frac{3m+4}{3}\right|\\ =\left\{\begin{array}{l}\left|3m+4\right|=6\\ \frac{\left|2m+1\right|}{2}\le \frac{\left|3m+4\right|}{3}\end{array}\right.\\ \iff m=\frac{2}{3}.\end{array}$